In Example 12-1 we studied the four different types of heat exchanger on an exothermic reaction. In this section we carry out the same study on an endothermic reaction.
Example 12-2. Production of Acetic Anhydride
Jeffreys,2 in a treatment of the design of an acetic anhydride manufacturing facility, states that one of the key steps is the endothermic vapor-phase cracking of acetone to ketene and methane:
CH3COCH3 → CH2CO CH4
He states further that this reaction is first-order with respect to acetone and that the specific reaction rate can be expressed by
where k is in reciprocal seconds and T is in Kelvin. In this design it is desired to feed 7850 kg of acetone per hour to a tubular reactor. The reactor consists of a bank of 1000 1-inch schedule 40 tubes. We shall consider four cases. The inlet temperature and pressure are the same for all cases at 1035 K and 162 kPa (1.6 atm) and the entering heating fluid temperature available is 1250 K.
A bank of 1000 1-in. schedule 40 tubes 1.79 m in length corresponds to 1.0 m3 (0.001 m3/tube = 1.0 dm3/tube) and gives 20% conversion. Ketene is unstable and tends to explode, which is a good reason to keep the conversion low. However, the pipe material and schedule size should be checked to learn if they are suitable for these temperatures and pressures. The heat exchange fluid has a flow rate, , of 0.111 mol/s, with a heat capacity of 34.5 J/mol·K.
Gas-phase endothermic reaction examples
Case 1 The reactor is operated adiabatically.
Case 2 Constant heat exchange fluid temperature Ta = 1250 K
Case 3 Co-current heat exchange with Ta0 = 1250 K
Case 4 Counter current heat exchange with Ta0 = 1250 K
Additional information Solution
Let A = CH3COCH3, B = CH2CO, and C = CH4. Rewriting the reaction symbolically gives us
A → B + C
Algorithm for a PFR with Heat Effects
Rearranging (E12-2.1)
Before combining Equations (E12-2.2) and (E12-2.6), it is first necessary to use the energy balance to determine T as a function of X.
a. Reactor balance
E12-2.7
b. Heat Exchanger. We will use the heat exchange fluid balance for co-current flow as our base case. We will then show how we can very easily modify our ODE solver program (e.g., Polymath) to solve for the other cases by simply multiplying the appropriate line in the code by either zero or minus one.
For co-current flow:
E12-2.8
Thermodynamics:
a. : At 298 K, using the standard heats of formation:
b. ΔCP : Using the mean heat capacities:
ΔCP = CPB – CPC – CPA = (83 + 71 – 163) J/mol · K
ΔCP = –9 J/mol · K
Heat Exchange:
Energy balance. From the adiabatic case in Case I, we already have CP, CPA. The heat-transfer area per unit volume of pipe is
Combining the overall heat-transfer coefficient with the area yields
Ua = 16,500 J/m3 · s · K
Table E12-2.1. Summary of Parameter Values
We will solve for all four cases for this endothermic reaction example in the same way we did for the exothermic reaction in Example 12-1. That is, we will write the Polymath equations for the case of co-current heat exchange and use that as the base case. We will then manipulate the different terms in the heat transfer fluid balance (Equations 12-10 and 12-11) to solve for the other cases, starting with the adiabatic case where we multiply the heat transfer coefficient in the base case by zero.
Case 1 Adiabatic
We are going to start with the adiabatic case first to show the dramatic effects of how the reaction dies out as the temperature drops. In fact, we are going to extend each to a volume of 5 dm3 to observe this effect and to show the necessity of adding a heat exchanger. For the adiabatic case, we simply multiply Ua in our Polymath Program by zero. No other changes are necessary. For the adiabatic case the answer will be the same whether we use a bank of 1000 reactors each 1 dm3 reactors or one of 1 m3. To illustrate how on endothermic reaction can virtually die out completely, let’s extend the single pipe volume from 1 dm3 to 5 dm3.
Ua = 16,500 * 0
The Polymath program is shown in Table E12-2.2. Figure E12-2.1 shows the graphical output.
Table E12-2.2. Polymath Program and Output for Adiabatic Operation
Figure E12-2.1. Adiabatic conversion and temperature (a) and reaction rate (b) profiles.
Analysis: Case 1 Adiabatic Operation: As temperature drops, so does k and hence the rate, –rA, drops to an insignificant value. Note that for this adiabatic endothermic reaction, the reaction virtually dies out after 3.5 dm3, owing to the large drop in temperature, and very little conversion is achieved beyond this point. One way to increase the conversion would be to add a diluent such as nitrogen, which could supply the sensible heat for this endothermic reaction. However, if too much diluent is added, the concentration, and hence the rate, will be quite low. On the other hand, if too little diluent is added, the temperature will drop and virtually extinguish the reaction. How much diluent to add is left as an exercise. While Figures E12-2.1 (a) and (b) give the reactor length as 5 dm3, the program and Table E12-2.2 are for 1 dm3 (0.001 m3) in order to easily compare the exit conversion with the other cases.
Case 2 Constant heat exchange fluid temperature, Ta
We make the following changes in our program on line 3 of the base case (a)
Table E12-2.3. Polymath Program and Output for Constant Ta
The profiles for T, X, and –rA are shown below.
Figure E12-2.2. Profiles for constant heat exchanger fluid temperature, Ta (a) temperature, (b) conversion, (c) reaction rate.
Analysis: Case 2 Constant Ta: Just after the reactor entrance, the reaction temperature drops as the sensible heat of reaction supplies the energy for the endothermic reaction. This temperature drop also causes the rate of reaction to drop. As we move further down the reactor, the reaction rate drops further as the reactants are consumed. Eventually down the reactor, the heat supplied by the constant Ta heat exchanger becomes greater than that “consumed” by the endothermic reaction and the reactor temperature rises. Consequently, we are eventually able to achieve an exit conversion of 95%.
Case 3 Co-Current Heat Exchange
The energy balance on a co-current exchanger is
with Ta0 = 1250 K at V = 0
Table E12-2.4. Polymath Program and Output for Co-current Exchange
The variable profiles for T, Ta, X, and –rA are shown below
Figure E12-2.3. Profiles down the reactor for an endothermic reaction with co-current heat exchange (a) temperature, (b) conversion, and (c) reaction rate.
Analysis: Case 3 Co-Current Exchange: In co-current heat exchange we see that the heat exchanger fluid temperature, Ta, drops rapidly initially and then continues to drop along the length of the reactor as it supplies the energy to the heat drawn by the endothermic reaction. Because the reactor temperature for co-current exchange is lower than that for constant Ta, the reaction rate will be lower. As a result, significantly less conversion will be achieved than in the case of constant heat exchange temperature Ta.
Case 4 Counter Current Heat Exchange
For counter current exchange we first multiply the rhs of the co-current heat exchanger energy balance by –1, leaving the rest of the Polymath program in Table 12-2.5 the same.
Next, guess Ta (V = 0) = 995.15 K to obtain Ta0 = 1250 K. (Don’t you believe for a moment 995.15 K was my first guess.) Once this match is obtained as shown in Table E12-2.5, we can report the profiles shown in Figure E12-2.4.
Table E12-2.5. Polymath Program and Output for Counter Current Exchange
Figure E12-2.4. Profiles down the reactor for counter current heat exchange (a) temperature, (b) conversion, (c) reaction rate.
Analysis: Case 4 Counter Current Exchange: One notes that the reactor temperature goes through a minimum along the length of the reactor. At the front of the reactor, the reaction takes place very rapidly, drawing energy from the sensible heat of the gas causing the gas temperature to drop because the heat exchanger cannot supply energy at an equal or greater rate. Additional “heat” is lost at the entrance in the case of counter current exchange because the temperature of the exchange fluid, Ta, is below the entering reactor temperature, T. This drop in temperature, coupled with the consumption of reactants, slows the reaction rate as we move down the reactor, resulting in a lower conversion than in any of the other three cases in this example. Near the middle of the reactor the rate slows down as the reactants become depleted and the heat exchanger now supplies energy at a rate greater than the reaction draws energy and as a result, the temperature eventually increases.
AspenTech: Example 12-2 has also been formulated in AspenTech and can be loaded on your computer directly from the DVD-ROM.
In this section we apply the general energy balance [Equation (11-22)] to the CSTR at steady state. We then present example problems showing how the mole and energy balances are combined to design reactors operating adiabatically and non-adiabatically.
In Chapter 11 the steady-state energy balance was derived:
Recall that is the shaft work, i.e., the work done by the stirrer or mixer in the CSTR on the reacting fluid inside the CSTR. Consequently, because the convention that done by the system on the surroundings is positive, the CSTR stirrer work will be a negative number, e.g., . [See problem P12-6B, a California Professional Engineers’ Exam Problem.]
[Note: In many calculations the CSTR mole balance derived in Chapter 2
(FA0X = –rAV)
will be used to replace the term following the brackets in Equation (11-28), that is, (FA0X) will be replaced by (–rAV) to arrive at Equation (12-12).]
Rearranging yields the steady-state energy balance
Although the CSTR is well mixed and the temperature is uniform throughout the reaction vessel, these conditions do not mean that the reaction is carried out isothermally. Isothermal operation occurs when the feed temperature is identical to the temperature of the fluid inside the CSTR.
The Term in the CSTR
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